exercise 7 - stemjock.com books/strauss pdes 2e/chapter 9... · 2019. 10. 11. · strauss pdes 2e:...

Strauss PDEs 2e: Section 9.2 - Exercise 7 of 12

Exercise 7

(a) Solve the wave equation in three dimensions for t > 0 with the initial conditions φ(x) = Afor |x| < ρ, φ(x) = 0 for |x| > ρ, and ψ|x| ≡ 0, where A is a constant. (This is somewhatlike the plucked string.) (Hint: Differentiate the solution in Exercise 6(b).)

(b) Sketch the regions in space-time that illustrate your answer. Where does the solution havejump discontinuities?

(c) Let |x0| < ρ. Ride the wave along a light ray emanating from (x0, 0). That is, look atu(x0 + tv, t) where |v| = c. Prove that

t · u(x0 + tv, t) converges as t→∞.

Solution

Part (a)

The solution of the three-dimensional wave equation in space subject to two initial conditions,

utt = c2∇2u, −∞ < x, y, z <∞, t > 0

u(x, y, z, 0) = α(x, y, z)

ut(x, y, z, 0) = β(x, y, z),

is given by the formula of Kirchhoff and Poisson.

u(x, y, z, t) =∂

∂t

[1

4πc2t

¨

(x0−x)2+(y0−y)2+(z0−z)2=c2t2

α(x0, y0, z0) dS0

]+

1

4πc2t

¨

(x0−x)2+(y0−y)2+(z0−z)2=c2t2

β(x0, y0, z0) dS0

In particular, we wish to solve the initial value problem when

α(x, y, z) =

{A r < ρ

0 r > ρand β(x, y, z) = 0,

where r = |x| =√x2 + y2 + z2. With these initial conditions, the previous formula simplifies to

u(x, y, z, t) =∂

∂t

(1

4πc2t

¨

Q∩T

AdS0

),

where

Q = {(x0, y0, z0) | (x0 − x)2 + (y0 − y)2 + (z0 − z)2 = c2t2}T = {(x0, y0, z0) | x20 + y20 + z20 < ρ2}.

Basically, this surface integral is over the part of the sphere centered at (x, y, z) with radius ctthat lies within the solid ball centered at the origin with radius ρ. Depending what region inspace-time the point (x, y, z, t) is chosen, the surface integral will yield a different result.

www.stemjock.com


The characteristic surfaces, which are obtained by drawing light rays (with slope c) from everypoint on the boundary of the hyperdisk within which the initial condition is nonzero, separatethese regions.

The red hyperdisk represents the solid ball in x0y0z0-space where the initial condition is nonzero.Because it has radius ρ, the height of the cone formed by the crossing characteristic lines is ρ/c. uwill be calculated within this cone first.

The First Region

The blue circle illustrated above represents the sphere centered at (x, y, z) with radius ct. Since itlies within the red disk, the surface area of the blue sphere that lies within the red ball is 4π(ct)2.Consequently,

u(x, y, z, t) =∂

∂t

(1

4πc2t

¨

Q∩T

AdS0

)

=∂

∂t

[A

4πc2t(4πc2t2)

]=

∂

∂t(At)

= A.

www.stemjock.com


This formula for u is valid at points in space-time where ρ > ct+ r, or r < ρ− ct. u will now becalculated in the region directly above the cone.

The Second Region

The blue circle illustrated above represents the sphere centered at (x, y, z) with radius ct. Sincethe red disk lies within it, the surface area of the blue sphere that lies within the red ball is zero.Consequently,

u(x, y, z, t) =∂

∂t

(1

4πc2t

¨

Q∩T

AdS0

)

=∂

∂t

[A

4πc2t(0)

]= 0.

This formula for u is valid at points in space-time where ct > ρ+ r, or r < ct− ρ. u will now becalculated in the region outside the cone right above the x0y0z0-plane.

The Third Region

The blue circle illustrated above represents the sphere centered at (x, y, z) with radius ct. Since itand the red disk are completely separate, the surface area of the blue sphere that lies within the

www.stemjock.com


red ball is zero. Consequently,

u(x, y, z, t) =∂

∂t

(1

4πc2t

¨

Q∩T

AdS0

)

=∂

∂t

[A

4πc2t(0)

]= 0.

This formula for u is valid at points in space-time where r > ρ+ ct. u will now be calculated inthe last region outside the cone.

The Fourth Region

The blue circle illustrated above represents the sphere centered at (x, y, z) with radius ct. Since itpartially intersects the red disk, the surface area of the portion of the blue sphere that lies withinthe red ball is

π(ct)

r[ρ2 − (r − ct)2].

Consequently,

u(x, y, z, t) =∂

∂t

(1

4πc2t

¨

Q∩T

AdS0

)

=∂

∂t

{A

4πc2t

π(ct)

r[ρ2 − (r − ct)2]

}=

∂

∂t

{A

4cr[ρ2 − (r − ct)2]

}=

A

4cr[−2(r − ct)(−c)]

=A

2r(r − ct).

This formula for u is valid at points in space-time where |ρ− ct| < r < ρ+ ct.

www.stemjock.com


To summarize the results, we have

u(x, y, z, t) =

A if r < ρ− ct0 if r < ct− ρ0 if r > ρ+ ctA

2r(r − ct) if |ρ− ct| < r < ρ+ ct

.

Therefore, replacing r with√x2 + y2 + z2,

u(x, y, z, t) =

A if√x2 + y2 + z2 < ρ− ct

0 if√x2 + y2 + z2 < ct− ρ

0 if√x2 + y2 + z2 > ρ+ ct

A

2√x2 + y2 + z2

(√x2 + y2 + z2 − ct) if |ρ− ct| <

√x2 + y2 + z2 < ρ+ ct

.

Part (b)

Space-time is illustrated below; the solution to the wave equation in each region is labeled bycolor.

Note that substituting ρ− ct, ct− ρ, and ρ+ ct for√x2 + y2 + z2 in the magenta solution does

not result in the blue, olive, and purple solutions, respectively. In other words, u is discontinuousacross each region.

www.stemjock.com


A movie of the solution will now be produced. Set ρ = c = A = 1 and replace√x2 + y2 + z2 with

|x| in the solution.

u(x, y, z, t) =

1 if |x| < 1− t0 if |x| < t− 1

0 if |x| > 1 + t1

2|x|(|x| − t) if |1− t| < |x| < 1 + t

u is only a function of |x| and t, so u = u(|x|, t). If t = 0, then

u(|x|, 0) =

1 if |x| < 1

0 if |x| < −10 if |x| > 11

2if 1 < |x| < 1

.

www.stemjock.com


If t = 14 , then

u

(|x|, 1

4

)=

1 if |x| < 3

4

0 if |x| < −3

4

0 if |x| > 5

41

2− 1

8|x|if3

4< |x| < 5

4

.

If t = 12 , then

u

(|x|, 1

2

)=

1 if |x| < 1

2

0 if |x| < −1

2

0 if |x| > 3

21

2− 1

4|x|if1

2< |x| < 3

2

.

www.stemjock.com


If t = 34 , then

u

(|x|, 3

4

)=

1 if |x| < 1

4

0 if |x| < −1

4

0 if |x| > 7

41

2− 3

8|x|if1

4< |x| < 7

4

.

If t = 0.95, then

u(|x|, 0.95) =

1 if |x| < 0.05

0 if |x| < −0.050 if |x| > 1.951

2− 0.95

2|x|if 0.05 < |x| < 1.95

.

www.stemjock.com


If t = 1.05, then

u(|x|, 1.05) =

1 if |x| < −0.050 if |x| < 0.05

0 if |x| > 2.051

2− 1.05

2|x|if 0.05 < |x| < 2.05

.

If t = 54 , then

u

(|x|, 5

4

)=

1 if |x| < −1

4

0 if |x| < 1

4

0 if |x| > 9

41

2− 5

8|x|if1

4< |x| < 9

4

.

www.stemjock.com


If t = 2, then

u(|x|, 2) =

1 if |x| < −10 if |x| < 1

0 if |x| > 31

2− 1

|x|if 1 < |x| < 3

.

www.stemjock.com


Plots of the amplitude will be shown for observers standing at |x| = 14 and |x| = 3

2 . If |x| =14 , then

u

(1

4, t

)=

1 if t <3

4

0 if t >5

4

0 if t < −3

41

2− 2t if

3

4< t <

5

4

.

If |x| = 32 , then

u

(3

2, t

)=

1 if t < −1

2

0 if t >5

2

0 if t <1

21

2− t

3if1

2< t <

5

2

.

www.stemjock.com


Part (c)

Let x1 be a point in space-time within the red hyperdisk initially: |x1| < ρ when t = 0. The aimhere is to show that

limt→∞

t · u(x1 + tv, t)

converges, where |v| = c. Replace√x2 + y2 + z2 with |x| and (x, y, z) with x in the solution for u.

u(x, t) =

A if |x| < ρ− ct0 if |x| < ct− ρ0 if |x| > ρ+ ctA

2|x|(|x| − ct) if |ρ− ct| < |x| < ρ+ ct

(x1 + tv, t) lies within the magenta region for large t, so

limt→∞

t · u(x1 + tv, t) = limt→∞

t · A

2|x1 + tv|(|x1 + tv| − ct)

= limt→∞

At

2√(x1 + tv)2

(√(x1 + tv)2 − ct)

= limt→∞

At

2√|x1|2 + 2tx1 · v + t2|v|2

(√|x1|2 + 2tx1 · v + t2|v|2 − ct)

= limt→∞

At

2√|x1|2 + 2tx1 · v + c2t2

(√|x1|2 + 2tx1 · v + c2t2 − ct)

= limt→∞

A2t

√|x1|2 + 2tx1 · v + c2t2

· ct(1

ct

√|x1|2 + 2tx1 · v + c2t2 − 1

)

= limt→∞

A

2√|x1|2t2

+ 2x1·vt + c2

· ct

(√|x1|2c2t2

+2x1 · vc2t

+ 1− 1

)

=

limt→∞

A

2

√��|x1|2t2 0

+��2x1·vt 0

+c2

limt→∞

ct

(√1 +

2x1 · vc2t

+|x1|2c2t2

− 1

).

Use the binomial series for the square root in the second limit.

=

(A

2c

)limt→∞

ct

[1 +

x1 · vc2t

+|x1|2

2c2t2+O

(1

t2

)− 1

]=A

2climt→∞

[x1 · vc

+��|x1|2

2ct0

+�

��O

(1

t

)0

]

=A

2c2x1 · v

Therefore, t · u(x1 + tv, t) converges as t→∞.

www.stemjock.com

exercise 7 - stemjock.com books/strauss pdes 2e/chapter 9... · 2019. 10. 11. · strauss pdes 2e:...

Documents